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According to https://stacks.math.columbia.edu/tag/08J5 we have for every complex $K$ and integer $a$ a distinguished triangle $$\tau_{\leq a}K\rightarrow\tau_{\leq a+1}K\rightarrow H^{a+1}(K)[-a-1]\rightarrow\tau_{\leq a}K[1].$$ If we consider the associated long exact sequence we get $$0=H^{a+1}(\tau_{\leq a}K)\rightarrow H^{a+1}(\tau_{\leq a+1}K)=H^{a+1}(K)\rightarrow H^{a+1}(H^{a+1}(K)[-a-1])=0$$ and so we should have $H^{a+1}(K)=0$, but there is no reason for this to be true. What is going wrong here?

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I'm a newb with derived categories, so this may be wrong:

What's the convention used for the shift functors in the stacks project? I think the confusion comes from the fact that we are using cochains in $K(\mathcal A)$.

In section 13.9, we are referred back to the following:

https://stacks.math.columbia.edu/tag/011G

So, if $A^\bullet$ is a cochain complex in $\mathcal A$, the shifted complex is

$A[k]^n = A^{n+k}$

So

$A^\bullet: \cdots \rightarrow A^{-2} \rightarrow A^{-1} \rightarrow A^0 \rightarrow A^1 \rightarrow A^2 \rightarrow \cdots $

gets shifted to

$A[k]^\bullet: \cdots \rightarrow A^{-2 + k} \rightarrow A^{-1 + k} \rightarrow A^k \rightarrow A^{1+k} \rightarrow A^{2+k} \rightarrow \cdots$

That is, $[1]$ is a left shift !

That makes sense because the $C \rightarrow A[1]$ part of a distinguished triangle should record the map $C^n \rightarrow A^{n+1}$ in degree $n$.

Now, I believe the convention is that $H^{a+1}(K^\bullet)$ denotes the complex

$ \cdots \rightarrow 0 \rightarrow 0 \rightarrow H^{a+1}(K^\bullet) \rightarrow 0 \rightarrow 0 \cdots$,

with $(H^{a+1}(K^\bullet))^0 = H^{a+1}(K^\bullet)$ and $H^{a+1}(K^\bullet))^n = 0$ otherwise.

Then, the right shift $H^{a+1}(K^\bullet)[-a-1]$ is the complex with

$(H^{a+1}(K^\bullet)[-a-1])^n = \begin{cases} H^{a+1}(K^\bullet), & n=a+1 \\ 0, & \mbox{otherwise} \end{cases}$

so then

$H^n(H^{a+1}(K^\bullet)[-a-1])) = \begin{cases} H^{a+1}(K^\bullet), & n=a+1 \\ 0, & \mbox{otherwise} \end{cases}$

and the long exact sequence associated to that distinguished triangle is

$\scriptsize \require{AMScd} \begin{CD} \cdots H^a(\tau_{\leq a}K^\bullet) @>>> H^a(\tau_{\leq a+1} K^\bullet) @>>> H^a(H^{a+1}(K^\bullet)) @>>> H^{a+1}(\tau_{\leq a}K^\bullet) @>>> H^{a+1}(\tau_{\leq a+1} K^\bullet) @>>> H^{a+1}(H^{a+1}(K^\bullet)) @>>> H^{a+2}(\tau_{\leq a} K^\bullet) \cdots \\ @VVV @VV=V @VV=V @VV=V @VV=V @VV=V @VV=V \\ \cdots H^a(K^\bullet) @>>> H^a(K^\bullet) @>>> 0 @>>> 0 @>>> H^{a+1}(K^\bullet) @>>> H^{a+1}(K^\bullet) @>>> 0 \cdots \end{CD}$

(Sorry if that goes off your screen, I'm not sure how to make arrows shorter in amscd)

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  • $\begingroup$ Indeed, you are right. Thanks! For some reason I thought it was a left shift.. $\endgroup$ – Bernie May 8 at 12:06

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